proofs
Provedattentioninnovationexternal-review-3

Surrogate innovation: exact mean-error bias algebra

For all J∈Type⁡\htmlData{mx=1}{J \in \htmlData{mx=0}{\operatorname{Type}}}, r,a,b,v:J→R\htmlData{mx=3}{r, a, b, v : \htmlData{mx=0}{J} \mathrel{\htmlData{mx=2}{\to}} \htmlData{mx=1}{\mathbb{R}}} and p,B,c,o,e∈R\htmlData{mx=1}{p, B, c, o, e \in \htmlData{mx=0}{\mathbb{R}}}, if p≠0\htmlData{mx=0}{p} \mathrel{\htmlData{mx=1}{\neq}} 0, ∑jr(j)⋅a(j)=p\htmlData{mx=7}{\sum_{j} \htmlData{mx=6}{\htmlData{mx=2}{\htmlData{mx=0}{r}(\htmlData{mx=1}{j})} \cdot \htmlData{mx=5}{\htmlData{mx=3}{a}(\htmlData{mx=4}{j})}}} \mathrel{\htmlData{mx=9}{=}} \htmlData{mx=8}{p}, ∑jr(j)⋅b(j)=B\htmlData{mx=7}{\sum_{j} \htmlData{mx=6}{\htmlData{mx=2}{\htmlData{mx=0}{r}(\htmlData{mx=1}{j})} \cdot \htmlData{mx=5}{\htmlData{mx=3}{b}(\htmlData{mx=4}{j})}}} \mathrel{\htmlData{mx=9}{=}} \htmlData{mx=8}{B} and ∑jr(j)⋅((a(j)−b(j))⋅(v(j)−c))=p⋅(o−c)\htmlData{mx=17}{\sum_{j} \htmlData{mx=16}{\htmlData{mx=2}{\htmlData{mx=0}{r}(\htmlData{mx=1}{j})} \cdot (\htmlData{mx=15}{(\htmlData{mx=9}{\htmlData{mx=5}{\htmlData{mx=3}{a}(\htmlData{mx=4}{j})} - \htmlData{mx=8}{\htmlData{mx=6}{b}(\htmlData{mx=7}{j})}}) \cdot (\htmlData{mx=14}{\htmlData{mx=12}{\htmlData{mx=10}{v}(\htmlData{mx=11}{j})} - \htmlData{mx=13}{c}})})}} \mathrel{\htmlData{mx=23}{=}} \htmlData{mx=22}{\htmlData{mx=18}{p} \cdot (\htmlData{mx=21}{\htmlData{mx=19}{o} - \htmlData{mx=20}{c}})} (given Fintype⁡ J\htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fintype}}\,\htmlData{mx=1}{J}}), then

c+e+∑jr(j)⋅((a(j)−b(j))⋅(v(j)−(c+e)))p−o=Bp⋅e\begin{aligned} & \htmlData{mx=28}{\htmlData{mx=26}{\htmlData{mx=3}{\htmlData{mx=1}{c} + \htmlData{mx=2}{e}} + \htmlData{mx=25}{\frac{\htmlData{mx=23}{\sum_{j} \htmlData{mx=22}{\htmlData{mx=6}{\htmlData{mx=4}{r}(\htmlData{mx=5}{j})} \cdot (\htmlData{mx=21}{(\htmlData{mx=13}{\htmlData{mx=9}{\htmlData{mx=7}{a}(\htmlData{mx=8}{j})} - \htmlData{mx=12}{\htmlData{mx=10}{b}(\htmlData{mx=11}{j})}}) \cdot (\htmlData{mx=20}{\htmlData{mx=16}{\htmlData{mx=14}{v}(\htmlData{mx=15}{j})} - (\htmlData{mx=19}{\htmlData{mx=17}{c} + \htmlData{mx=18}{e}})})})}}}{\htmlData{mx=24}{p}}}} - \htmlData{mx=27}{o}} \\ & \quad \htmlData{mx=0}{=} \htmlData{mx=33}{\htmlData{mx=31}{\frac{\htmlData{mx=29}{B}}{\htmlData{mx=30}{p}}} \cdot \htmlData{mx=32}{e}} \end{aligned}
What the notation means · or point at any symbol
Type
Type: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (Type)
Fintype
Fintype: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (Fintype)
f(x)Function application
Fintype applied to J. Lean writes f x; papers write f(x). (Fintype J)
⟹Implies
If J, then ℝ. (Between types, A → B is the type of functions from A to B.) (J → ℝ)
x : TVariables
r, a, b, v range over J → ℝ: the statement must hold for every choice. ((r a b v : J → ℝ))
x ∈ TVariables
p, B, c, o, e range over real numbers: the statement must hold for every choice. ((p B c o e : ℝ))
≠Not equal
p and 0 are different. (p ≠ 0)
∑Summation
Add up r(j) · a(j) for every j. (∑ j, r j * a j)
=Equality
∑ j, r(j) · a(j) and p are the same. (∑ j, r j * a j = p)
+Addition
c plus e. (c + e)
RReal numbers
The real numbers: every point on the number line. (ℝ)
·Multiplication
r(j) times a(j). (r j * a j)
a/bDivision
∑ j, r(j) · ((a(j) − b(j)) · (v(j) − (c + e))) divided by p. In Lean, dividing by 0 gives 0. ((∑ j, r j * ((a j - b j) * (v j - (c + …)
-Subtraction
a(j) minus b(j). On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (a j - b j)

Finite scalar aggregate identity giving B/p times a surrogate mean error. The infinite-horizon expectation argument is a separate paper proof.

How it was proved

Proved
6 October 2026
By
Ajax, with the agent “Baguette math & proofs (Claude Code, 2026-10-03)”
Checked by
Lean's kernel (Lean 4.33.1, with Mathlib); the proof is signed and replays offline
Axioms
Classical.choice, Quot.sound, propext
Proof
15 lines of Lean
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